Leetcode_candy

地址：http://oj.leetcode.com/problems/candy/

There are N children standing in a line. Each child is assigned a rating value.

You are giving candies to these children subjected to the following requirements:

• Each child must have at least one candy.
• Children with a higher rating get more candies than their neighbors.

What is the minimum candies you must give?

思路：两次遍历，一次从头到尾，一次从尾到头，O(N)时间复杂度。

用分治法在第二十个case时wa了，case太多又无法自测。把wa的分治法代码贴出来，有哪位大哥大姐能看出错误请留个言啊。。。

参考代码：

class Solution {public:    int candy(vector<int> &ratings) {        if(ratings.empty())            return 0;        vector<int>vec(ratings.size(), 1);        for(int i = 0; i<ratings.size()-1; ++i)        {            if(ratings[i]<ratings[i+1])            {                vec[i+1] = vec[i]+1;            }        }        for(int i = ratings.size()-2; i>=0; --i)        {            if(ratings[i]>ratings[i+1])            {                vec[i] = max(vec[i], vec[i+1]+1);            }        }        return accumulate(vec.begin(), vec.end(), 0);    }};

分治法WA 代码：

class Solution {public:    void dispatch(vector< int>&ratings, vector< int>&vec, int st, int ed)    {        if(st < ed - 1)        {            int mid = (st+ed)/2;            dispatch(ratings, vec, st, mid);            dispatch(ratings, vec, mid+1, ed);            if(ratings[mid]<ratings[mid+1])            {                if(vec[mid]>=vec[mid+1])                {                    vec[mid+1] = vec[mid]+1;                    int i = mid+1;                    while(i<ed && ratings[i]<ratings[i+1])                    {                        vec[i+1] = vec[i]+1;                        ++i;                    }                }            }            else if(ratings[mid]>ratings[mid+1])            {                if(vec[mid]<=vec[mid+1])                {                    vec[mid] = vec[mid+1]+1;                    int i = mid;                    while(i>st && ratings[i]<ratings[i-1])                    {                        vec[i-1] = vec[i]+1;                        --i;                    }                }            }        }        else if(st==ed)            vec[st] = 1;        else        {            if(ratings[st]<ratings[ed])            {                vec[st] = 1;                vec[ed] = 2;            }            else if(ratings[st]>ratings[ed])            {                vec[st] = 2;                vec[ed] = 1;            }            else                vec[st] = vec[ed] = 1;        }    }    int candy(vector< int> &ratings) {        if(ratings.empty())            return 0;        vector< int>vec(ratings.size(), 0);        dispatch(ratings, vec, 0, ratings.size()-1);        int ans = accumulate(vec.begin(), vec.end(), 0);        return ans;    }};

//SECOND TRIAL, almost the same
class Solution {
public:
    int candy(vector<int> &ratings) {
        if(ratings.empty())
            return 0;
        vector<int>vec(ratings.size(), 0);
        vec[0] = vec[vec.size()-1] = 1;
        for(int i=0; i<ratings.size()-1; ++i)
            if(ratings[i]<ratings[i+1])
                vec[i+1] = vec[i]+1;
            else
                vec[i+1] = 1;
        for(int i=ratings.size()-2; i>=0; --i)
            if(ratings[i]>ratings[i+1])
                vec[i] = max(vec[i], 1+vec[i+1]);

        return accumulate(vec.begin(), vec.end(), 0);
    }
};

python version

class Solution:
    # @param ratings, a list of integer
    # @return an integer
    def candy(self, ratings):
        if not ratings:
            return 0
        vec = [1]*len(ratings)
        for i in range(0, len(ratings)-1):
            if ratings[i] < ratings[i+1]:
                vec[i+1] = 1 + vec[i]
        
        for i in range(len(ratings)-2, -1, -1):
            if ratings[i] > ratings[i+1]:
                vec[i] = max(vec[i], 1+vec[i+1])
        
        return sum(vec)